Jacoboson 4.2 Work Problem

2.1 Show that the regular pentagon can be constructed with straight-edge and compass.
Solution: Let \(z=e^{\frac{2\pi i}{5}}\),

\(x_1=z+z^4=2\cos{\frac{2\pi}{5}},\ x_2=z^2+z^3=2\cos{\frac{4\pi}{5}}.\)
Since \(z^5=1,z\neq 1\), we knew that
\(x_1+x_2=z+z^2+z^3+z^4=-1\), and \(x_1x_2=z^3+z^4+z+z^2=-1.\)
By vieta theorem, \(x_1,x_2\) are roots of the equation \(x^2+x-1=0\). We have
\(x_1=\frac{-1+\sqrt{5}}{2}\) and \(x_2=\frac{-1-\sqrt{5}}{2}\).
Thereby, \(x_1,x_2\in\mathbb{Q}(\sqrt{5}).\)
It is also easy to see that
\(z*z^4=z^5=1\), and \(z+z^4=x_1\). Thus \(z\) is a root of the equation \(y^2-x_1y+1=0\), and it is contained in the field \(\mathbb{Q}(\sqrt{5})(\sqrt{x_1^2-4}).\) Hence \(z\) is constructible.
Remark: 第一次看basic algebra 4.2节的时候我个人感觉这部分是很迷惑的，因为这个root tower的构造以及想法我认为并不明确。这个方法事实上是通过在\(\mathbb{Q}\)关于\(x^n-1\)的Galois extension通过galois pairing转到group的cases 发现一个链并再对应回field的cases从而得到回答中\(x_1,x_2\)的形式。这个链具体长什么样子可以去书里找找，这里打出来会比较累和麻烦，因为某些符号还没解释.

2.2 Show that \(\arccos{\frac{11}{16}}\) can be trisected with straight-edge and compass.
Solution: We need to prove that the number \(z=\cos{(\arccos{\frac{11}{16}})}\) is constructible. We have \(\cos{3x}\) equals to
\(\cos{2x}\cos{x}-sin{2x}\sin{x}=(2\cos^2{x}-1)\cos{x}=2\cos{x}(1-\cos^2{x})=4\cos^3{x}-3\cos{x}.\)
Hence \(z\) is a root of the equation \(4x^3-3x-\frac{11}{16}=0\). We have
\(4x^3-3x-\frac{11}{16}=\frac{1}{16}(x^3-12x-11)=\frac{1}{16}(x+1)(x^2-x-11).\)
Hence, \(z=\frac{1+3\sqrt{5}}{8}\) is constructible.

2.3 Show that the regular 9-gon cannot be constructed.
Solution: The complex number \(z=e^{\frac{2\pi i}{9}}\) is a root of \(x^9-1=0\). We factorize it into
\((x-1)(x^2+x+1)(x^6+x^3+1).\)
Thus \(x^6+x^3+1\) is the minimum polynomial of \(z\) over \(\mathbb{Q}\), and it is not the power of 2. Hence z is not constructible.